Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

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In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in JAVA):

Solution:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.PriorityQueue;
import java.util.Random;
import java.util.StringTokenizer;

public class a {

	public static void main(String[] args) throws IOException {
		FastScanner in = new FastScanner(System.in);
		PrintWriter out = new PrintWriter(System.out);
		int n = in.nextInt();
		int m = in.nextInt();
		int[][] arr = new int[n][m];
		for (int i = 0; i < n; i++) {
			for (int j = 0; j < m; j++) {
				arr[i][j] = in.nextInt();
			}
		}
		
		PriorityQueue<Range> pq = new PriorityQueue<>();
		for (int i = 0; i < n; i++) {
			pq.add(new Range(i, 0, i, m - 1, "row " + (i + 1)));
		}
		
		for (int i = 0; i < m; i++) {
			pq.add(new Range(0, i, n - 1, i, "col " + (i + 1)));
		}
		
		ArrayList<String> ans = new ArrayList<>();
		while (!pq.isEmpty()) {
			Range range = pq.poll();
			int min = 1000000;
			for (int r = range.r0; r <= range.r1; r++)
				for (int c = range.c0; c <= range.c1; c++)
					min = Math.min(min, arr[r][c]);
			if (min == 0) continue;
			for (int r = range.r0; r <= range.r1; r++)
				for (int c = range.c0; c <= range.c1; c++)
					arr[r][c] -= min;
			for (int i = 0; i < min; i++)
				ans.add(range.str);
		}
		int sum = 0;
		for (int i = 0; i < n; i++)
			for (int j = 0; j < m; j++)
				sum += arr[i][j];
		if (sum > 0) {
			out.println(-1);
		} else {
			out.println(ans.size());
			for (String str : ans)
				out.println(str);
		}
		out.close();
	}
	
	static class Range implements Comparable<Range> {
		int r0, c0, r1, c1, a;
		String str;
		public Range(int r00, int c00, int r11, int c11, String strr) {
			r0 = r00; c0 = c00; r1 = r11; c1 = c11;
			a = r1 + c1 - r0 - c0 + 1;
			str = strr;
		}
		public int compareTo(Range r) {
			return Integer.compare(r.a, a);
		}
	}
	
	//@
	static class FastScanner {
	    BufferedReader br;
	    StringTokenizer st;
		
	    public FastScanner(InputStream i) {
	        br = new BufferedReader(new InputStreamReader(i));
	        st = new StringTokenizer("");
	    }
				
	    public String next() throws IOException {
	        if(st.hasMoreTokens())
	            return st.nextToken();
	        else
	            st = new StringTokenizer(br.readLine());
	        return next();
	    }

	    public int nextInt() throws IOException {
	        return Integer.parseInt(next());
	    }
	    //#
	    public long nextLong() throws IOException {
	        return Long.parseLong(next());
	    }
	    public double nextDouble() throws IOException {
	        return Double.parseDouble(next());
	    }
	    //$
	}
	static class ArrayUtils {
		static Random rand = new Random();
		public static void sort(byte[] arr) {
			int n = arr.length;
			for (int i = 0; i < n; i++) {
				int x = rand.nextInt(n);
				byte tmp = arr[x];
				arr[x] = arr[i];
				arr[i] = tmp;
			}
			Arrays.sort(arr);
		}
		public static void sort(int[] arr) {
			int n = arr.length;
			for (int i = 0; i < n; i++) {
				int x = rand.nextInt(n);
				int tmp = arr[x];
				arr[x] = arr[i];
				arr[i] = tmp;
			}
			Arrays.sort(arr);
		}
		public static void sort(long[] arr) {
			int n = arr.length;
			for (int i = 0; i < n; i++) {
				int x = rand.nextInt(n);
				long tmp = arr[x];
				arr[x] = arr[i];
				arr[i] = tmp;
			}
			Arrays.sort(arr);
		}
		public static void sort(double[] arr) {
			int n = arr.length;
			for (int i = 0; i < n; i++) {
				int x = rand.nextInt(n);
				double tmp = arr[x];
				arr[x] = arr[i];
				arr[i] = tmp;
			}
			Arrays.sort(arr);
		}
	}
}